A boat crosses a river of width 244 m in which the current has a uniform speed of 1.99 m/s. The pilot maintains a bearing (i.e.,
the direction in which the boat points) perpendicular to the river and a throttle setting to give a constant speed of 1.68 m/s relative to the water. How far downstream from the initial position is the boat when it reaches the opposite shore
Since no friction present, assuming no external forces acting during the three collisions, total momentum must be conserved.
For the first collission, only mass 1 is moving before it, so we can write the following equation:
Since both masses are identical, and they stick together after the collision, we can express the final momentum as follows:
From (1) and (2) we get:
v₁ = v₀/2 (3)
Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1 and 2 combined together before colliding with mass 3 is just v₁, so the initial momentum prior the second collision (p₁) can be expressed as follows:
Since after the collision the three masses stick together, we can express this final momentum (p₂) as follows:
From (4) and (5) we get:
v₂ = v₀/3 (6)
Since the masses are moving on a frictionless 1D track, the speed of the set of mass 1, 2 and 3 combined together before colliding with mass 4 is just v₂, so the initial momentum prior the third collision (p₂) can be expressed as follows:
Since after the collision the four masses stick together, we can express this final momentum (p₃) as follows:
From (7) and (8) we get:
v₃ = v₀/4
This means that after the last collision, the speed will have been reduced to a 25% of the initial value, so it will have been reduced in a 75% regarding the initial value of v₀.
